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Thought blocker's explanation is correct but the proportionality of P to V2 wasn't mentioned so to just highlight that i still intended to post this ....http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
can someone help question 6 b please!
Np
I am online from tab, and it will take a lot time to reply, and I have to finish M1 so please help her asap
17 would be A cause if you complete the lines of the force acting all should intersect at the same point
Feels honored
Genius
thanks a lot help me can you please help me with other questions?
Please give me some time. tomorrow i have to go to college early so i will solve them for you when i come back.
thank you very much but this doesnt contain applications of physics, telecomunications , medical physics ...
Q25: The image of a wave has been given, and so has the direction it is traveling in.
Note that since this is a transverse wave and the disturbances are traveling from left to right, the particles at any point are vibrating in a direction perpendicular to that propagation.
In other words, if you focused on any one segment of the rope and that one segment alone (let's say by painting it a different color from that of the rope), then you would find it does not experience any sideways displacement; it only moves up, then down, then up, then down, etc.
At the same time, the waveform shifts towards the right, since that is the direction in which it is traveling.
So just imagine the waveform displayed in the question shift to the right; imagine each part of the waveform progress, and you would then see that P is moving downwards; since the trough just before P has to travel to the right, P has to "fall" into that trough, and if it "falls" into that trough, it is moving downwards.
Of course, this eliminates all the options in the question, leaving behind only A, the correct answer, but it is worth discussing the motion of Q:
At the instant of the displayed image, Q is a segment at the crest of the waveform; it is debatable that after some time, Q has to fall towards the equilibrium position, but taking a closer look at the wave motion (and seeing that we can estimate the motion of any one segment as Simple Harmonic Motion, SHM) it turns out that Q would be stationary.
It's just like the motion of a pendulum; at any extreme, the rate of change of it's position - it's velocity - is zero, but due to the forces acting on the pendulum (in THIS case, the force is the tension in the string) this changes slowly to make it change position. So that's most likely why the velocity of Q is zero at the instant shown.
Q26: There are two formulae we will need to use here; one relates Intensity of a wave to it's Amplitude, the other relates the Power of a wave to it's Intensity.
First, let's fix up an equation for the initial state/ situation (I = Intensity of Wave , A = Amplitude of Wave , k = arbitrary constant):
I = k * A^2
The other equation tells us (P = Power of Wave , S = Area component perpendicular to wave propagation direction):
P = I * S
Substituting the value of I from the first equation into the second,
P = k * A^2 * S
So all we have to do now is form two equations (one for each situation) and compare the two.
Initially,
P(i) = k * A^2 * S
Finally,
P(f) = k *4A^2 * 0.5S
Dividing the second equation by the first,
P(f) / P(i) = 2
Since P(i) = E,
P(f) = 2E =B
Q33: Seriously, this one is a doozy
What it actually asks for (as far as I can tell) are the variables required to find the different between OPEN circuit voltage and CLOSED circuit voltage. In other words, the difference between the potential difference measured when the circuit is OPEN and when it is fully connected with an external resistor.
(I thought it spoke about the normal decrease in voltage with running duration, which can be expected as a battery runs down. My mistake!)
A Google Images search for "internal external resistance" provides some very clear images, and i'm using the attached one for reference.
View attachment 38354
When the circuit is open, no current flows through the battery, and the initial reading the equivalent to the EMF of the battery; no energy is lost per second to heat in the internal resistance, so it is equal to the EMF.
Let's write this down as V.
Once the circuit is connected as above, a current indeed does flow through all components, and a potential difference is maintained across each too. However, since the battery consists of an internal resistance (this internal resistance is NEVER separate from the battery; it can be considered as such, but the battery consists both of the cells and the internal resistance), there is a drop in potential across this internal resistance before the current even leaves the battery. Thus, the potential difference across the terminals of the battery drop.
Applying Kirchoff's Second Law while moving from B to A, what we get is (V is still the EMF, r(internal) is the internal resistance):
(Potential at A) + V - i * r(internal) = (Potential at B)
All this says is that the potential at A is one thing, it changes by so and so amount, as a result of which is becomes another value, the Potential at B.
So, since the Potential Difference is the (Potential at B) - (Potential at A), we have:
PD = V - i * r(internal)
Thus, the drop in potential is
V - (V- i * r(internal)) = i * r(internal)
The change is negative, but the decrease is positive, so the signs change there. Therefore, all you need to find the drop in potential are the final current, and the internal resistance of the battery = C
Q35: Assuming that the ammeter and the voltmeter are both ideal, the resistance of the ammeter is Zero and the resistance of the voltmeter is Infinity. Following on from there, the potential difference across the ammeter is Zero and the current through the Voltmeter is Zero.
So these don not affect the circuit in any way except to give us a reading. When the variable resistance is dropped, applying Kirchoff's Second Law gives us:
i = V/(∑R) where ∑R is the equivalent resistance of the circuit, in this case the algebraic sum of the two resistance values.
Since the overall resistance of the circuit decreases, the current automatically decreases and the answers are narrowed down to C or D.
Furthermore, since we have
V = I * R
and the current across the fixed resistor increases while the resistance remains the same,
V(f) = I(f) * R
Dividing the second equation by the first, we get V(f) = [I(f) / I] * V.
Since [I(f) / I] > 1, V(f) > V and so the reading on the voltmeter also increases, telling us the answer is D.
Note: If the voltmeter was across the Variable Resistor, our calculation would have been a little longer, since the increase in Current and the decrease in Resistance might *possibly* make up for each other and result in Zero Change in the voltmeter reading.
Hope this helped!
Good luck for all your exams!
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